word_count challenge

Question

need a little help here guys.

I got stuck in this challenge for hours, and still cant find any solution. Anyone?

Attached is my latest code. Thanks guys!

word_count.py

# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.

some_dict = {'a': 1, 'b': 2, 'c': 3}

key_list = ['a', 'b', 'd']

def word_count(some_dict, key_list):
  counts = 0
  for item in key_list:
    if item in some_dict.keys():
      counts += 1
  return counts

Answer 1 · 2016-07-12T22:05:37Z

July 12, 2016 10:05pm

Here's a few hints:

you only need to create the function, the challenge will provide the test data
your function will take only one argument, a string
you will probably want to split the string to help loop through the words
your function will build a new dictionary
the counts will be kept in the new dictionary
you don't need the keys function to check if an item is in the dictionary
when the item is not already in the dictionary, a new item with a count of 1 will be added to it
remember to return the finished dictionary

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gibran erlangga

gibran erlangga

word_count challenge

1 Answer

Steven Parker

Steven Parker

Here's a few hints: